Introduction of an Elementary Method to Express ζ(2n+1) in Terms of ζ(2k) with k 1

Transcription

1 Itroductio of a Elemetary Method to Express ζ+ i Terms of ζk with k arxiv:85.3v [math-ph] 3 May 8 Kazuyuki FUJII ad Tatsuo SUZUKI Departmet of Mathematical Scieces Yokohama City Uiversity Yokohama, 36 7 Japa Ceter for Educatioal Assistace Shibaura Istitute of Techoy Saitama, Japa Abstract I this ote we give the most elemetary method as far as we kow to express ζ+ i terms of ζk k. The method is based o oly some elemetary works by Leohard Euler, so it is very istructive to o experts or studets. address : fujii@yokohama-cu.ac.jp address : suzukita@aoi.waseda.jp ; i7@sic.shibaura-it.ac.jp

2 The zeta fuctio ζx is defied by ζx = x for x > ad we are iterested i the zeta values ζk k ζk = k. The value of ζk is well kow by Euler ζ = π 6 π4, ζ4 =, etc. 9, while that of ζk + is less kow except for ζ3 is irratioal by Apéry. See [] ad its refereces. The textbook [] is also recommeded. Therefore it is desirable to express ζk+ i terms of ζk k like ζk+ = c + c ζ 3 where c are costats. I this ote we revisit the problem ad give a method by use of oly a few elemetary works by Euler. It is most elemetary as far as we kow. Let us start by listig well kow works by Euler : a six = eix e ix i b six = x x π c xsixdx = π ζ3 Though the equatio c may be ot popular amog the Euler s works which are huge! [3] it is importat ad iterestig eough as show i the followig. For the sake of o experts who are iterested i the zeta values we show our method with simple example c, which is very istructive. Namely, we calculate the itegral we do t kow the precise defiitio of most elemetary x si xdx 4

3 i two ways by use of a ad b. Here we list some well kow itegrals related to 4 for the coveiece of readers. sixdx = π sixsixdx = xxdx = π π 8 π 6 I Calculatio by use of a By a six = eix e ix = e ix e ix i = e ix e ix i i = ix i+ e ix e ix = ix i, 5 where we have used the Taylor expasio z z = for z <. The xsixdx = i x dx i = i π3 4 π 8 i xdx = xe ix dx xe ix dx ad xe ix dx = [ ]π e ix i x + e ix dx i = iπ 4 e iπ 4 e iπ = iπ

4 Therefore Sice it is easy to see ad we obtai xsixdx = i π3 4 π 8 i = = i π3 4 π 8 i iπ 4 = i π3 4 π 8 i+ iπ 4 = 3 iπ = = 7 8 ζ = ζ xsixdx = i π3 4 π 8 i+ iπ 4 ζ+ 7 8 ζ3 where we have used the pricipal value = i π3 4 π 8 +i+ iπ 8 ζ+ 7 6 ζ3 = π π 3 6 ζ3+i 4 π3 6 + π 8 ζ = π ζ 6 ζ3+iπ π, i = Loge iπ = i π. As a result we have ζ = π 6 automatically ad the equatio c. A commet is i order. Our proof is ot rigorous i the mathematical sese because the covergece radius of 5 is igored. O the other had, it is very clear why ζ3 appears i the process of calculatio. II Calculatio by use of b 4

5 so By b xsixdx = six = x+ xxdx+ x, 7 π = π π 8 π 6 + x x π dx x x π Let us calculate the last term. By the chage of variables x π x ad usig the Taylor expasio we obtai x As a result we have x dx x = π 4 xdx π x = dx x = π π = π x 4 x k dx k xsixdx = π π 8 π 6 π 8 = π π 8 π 6 π 8 = kk + 4 k+ π k+ 8 dx. kk + k k. kk + k k ζk kk +k. 8 By comparig I with II we obtai the expressio of ζ3 ζ3 = π π 7 ζ + However, our expressio 9 is of course ot ew, see for example [4] or [5]. By the way, the expressio by Euler is differet from ours : ζ3 = π ζ ,

6 see []. Therefore these two expressios give the iterestig equatio π = + ζ π + = e ζ +. I the followig we geeralize the method above to obtai the equatio3. For that purpose we cosider the itegral It may be reasoable to call this the Euler itegral. We calculate i two ways by use of a ad b. x l sixdx for l. I Calculatio by use of a I a similar way i I it is easy to see x l sixdx = i x l dx i = i π l+ l + +i π x l dx π l l = i π l+ l + iπ l+ π l l = i π l+ ll+ π l l x l e ix dx x l e ix dx x l e ix dx x l e ix dx. I order to calculate the last term which is ot so easy we make use of the trick. From e ix dx = e iπ i = i e iπ we differetiate the equatio above l times with respect to i l = i l j= = l! i l d e iπ d x l e ix dx = i l! j!l j! iπl j e iπ j j! j +e iπ l l! l l j= iπ l j + l j! j+ l 6

7 where we have used the Leibiz s rule of differetiatio, so l x l e ix dx = l! i l j= = l l! l iπ l j l j! j+ l j= + l iπ l j + l j! j+ l. By otig = ζk, k k = ζk k k we obtai 7

8 = i = i x l sixdx π l+ ll + π l l π l+ ll + π l l + l l! l + l l! l l iπ l j l j! j= j+ + l+ l iπ l j ζj ++ ζl+ l j! j+ l+ j= by dividig the summatio ito j = k k =,,l ad j = k k =,,l = i = π l l +i π l+ ll + π l l = π l l +i + l l! l + l l! l l k= l + l l! l iπ l k l k! iπ l k l k! l ζk+ k+ k l k π l k l k! + l l! l+ 4l ζl+ π l+ ll + + l l! l + l! l l k π l k l k! l l k π l k l k! k= k + l l! l+ ζl+ 4l π l+ l! + ll + l l k= k π l k l k! ζk++ l+ ζk+ k ζk+ k+ From this equatio the imagiary part must be zero, so we have l! l l k= k π l k ζk+ π l+ l k! k+ ll+ 8 l ζl+ ζk+ k+ ζk+. = l =,,.

9 Here, if we asuume ζ = 3 the the equatio above is rewritte i a compact form or k k l k= l k= k π l k ζk+ = l =,, l k! k+ k π l k+ l k+! From this we have the recurret relatio Result For l =,, ζl = l l Let us list some examples : l k= ζk = l =,,. k l+k π l k l k+! ζ = π π4, ζ4 = 6 9, etc. Next, from the real part of the equatio we have Result For l =,, x l sixdx = π l l Let us list some examples : + l! l l ζk. 4 k k π l k l k! ζk+ k + l l! l+ 4l ζl+. 5 xsixdx = π ζ3, x 3 sixdx = π4 64 9π 93 + ζ3 ζ5, etc II Calculatio by use of b 9

10 I a similar way i II it is easy to see Therefore we have x l sixdx = = π l l = π l l = π l l = π l l = π l l π l = Result 3 For l =,, π π l x l sixdx = x l xdx+ π π l l + π l π π l l π l π π l l π l π π l l l π l π π l π l l l x l x π l 4 π l π l By comparig 5 with 6 we have the mai result Result 4 Mai For l =,, For examples, ζl+ = l l l+ ζ3 = π 7 ζ5 = 6π 3 = 4π4 65 l πl l! π k π l k l k! π l l dx t l tdt 4 t k+l dt k kk +l4 k+l kk +l k+l k l ζk kk +l k ζk kk +l k ζk+ k ζk, kk + k ζk kk +l k ζ3 π π 9 4 ζk kk + k π 9 8 k + ζk. kk +k + k ζk. 6 kk +l k.. 7

11 Acommet isiorder. There aremay expressios like 7. Forexample, i[6] Theorem A it is give ζl+ = l π l l l+ [ l ] k kζk+ π k l k! + ζkk!. k k +l! However, our expressio is differet from this. I this ote we gave a simple method to obtai some deep relatios amog zeta values by calculatig the Euler itegral i our termioy. The method is systematic ad most elemetary as far as we kow. Moreover, it must be fresh for ot oly o experts or studets but also experts i Elemetary Number Theory. We coclude this ote by statig our dream. We are workig i some field of Quatum Computatio, so we dream that the Riema cojecture will be fiished by it. Refereces [] WolframMathWorld : Riema Zeta Fuctio, [] M. Koecher : Klassische elemetare Aalysis, Birkhäuser Basel Bosto, 987. [3] W. Duham : Euler : The Master of Us All, Mathematical Ass of Amer, 999. [4] M. Sugioka : sugi m/page.htm [5] I. Sato : kotaro/idex.htm [6] D. Cvijović ad J. Kliowski : New Rapidly Coverget Series Represetatios For ζ +, Proc. Amer. Math. Soc., 5 997, 63 7.